Reversed limit definition for monotonic functions

[ecoo]

Homework Statement

Does the delta-epsilon limit definition in reverse work for describing limits in monotonic functions?

By reversed, one means for

lim (x -> a) f(x) = L

if for each δ there corresponds ε such that

0 < | x-a | < δ whenever | f(x) - L | < ε.

Homework Equations

The Attempt at a Solution

I am thinking that it works, because this definition means that the range interval must lie within the domain interval, and it can be seen that shrinking δ also shrinks ε, which is how the usual definition works but in reverse.

I don't think this would work for non-monotonic functions because there can be many f(x) that satisfy
| f(x) - L | < ε but not | f(x) - L | < ε. Hopefully someone can also confirm this part too.

Thanks

 

[fresh_42]

You can define whatever you like, as long as it is well defined. For this reason it should be helpful to indicate which variable depends on which and to explicitly write the quantors. If I read it correctly, then $\varepsilon = \varepsilon(\delta)$. But then I have difficulties with the next line. I read $\forall \, \delta \, \exists \, \varepsilon(\delta)\, : \,\vert \, f(x)-L\,\vert \, < \varepsilon(\delta) \Longrightarrow \,\vert \,x-a\,\vert \,< \delta \,$. The order of the quantors don't seem to reflect to the order of the conclusion. And how do you specify $L$\,?

The remaining question is in any case: What for?

 

[ecoo]

You can define whatever you like, as long as it is well defined. For this reason it should be helpful to indicate which variable depends on which and to explicitly write the quantors. If I read it correctly, then $\varepsilon = \varepsilon(\delta)$. But then I have difficulties with the next line. I read $\forall \, \delta \, \exists \, \varepsilon(\delta)\, : \,\vert \, f(x)-L\,\vert \, < \varepsilon(\delta) \Longrightarrow \,\vert \,x-a\,\vert \,< \delta \,$. The order of the quantors don't seem to reflect to the order of the conclusion. And how do you specify $L$\,?

I forgot the part where δ > 0 and ε > 0, so I think it would be written like this:

$\forall \, \delta \ > 0, \exists \, \varepsilon(\delta)\ > 0, : \,\vert \, f(x)-L\,\vert \, < \varepsilon(\delta) \Longrightarrow \, 0 < \vert \,x-a\,\vert \,< \delta \,$

L is the supposed limit.

lim (x -> a) f(x) = L

The remaining question is in any case: What for?

I'm not very experienced in this type of math, so it's just a random inquiry.

 

[fresh_42]

It doesn't really make sense. You can have $f(x) = L$ for several points $x$ which can be far away from $x=a$. That's where "what for" is needed. The way it is written now, is only confusing (IMO). It is more a condition of monotone behavior than on limits, because it says, $\lim_{x \to a}f(x)=L$ can only happen in a neighborhood of $x=a$ and nowhere else.